indefinite integrals

**macca101** · June 30th 2006, 12:04 PM

I'm working through my exercise book quite happily finding indefinite integrals using the table of indefinite integrals, I have come across this one and have been staring at it for an hour unable to understand how this answer is derived.

$ \int \frac{-6}{x^4} dx $

I know the answer is

$ \frac{2}{x^3} +c $

**CaptainBlack** · June 30th 2006, 12:11 PM

Originally Posted by macca101

I'm working through my exercise book quite happily finding indefinite integrals using the table of indefinite integrals, I have come across this one and have been staring at it for an hour unable to understand how this answer is derived.

$ \int \frac{-6}{x^4} dx $

I know the answer is

$ \frac{2}{x^3} +c $

Bog standard power integral:

$ \int \frac{-6}{x^4} dx=-6 \int x^{-4} dx=(-6)(-1/3)x^{-3}+c= \frac{2}{x^3} +c $

RonL

**macca101** · June 30th 2006, 12:46 PM

Thanks

Thread: indefinite integrals

LinkBack

Thread Tools

Display

indefinite integrals

Similar Math Help Forum Discussions

Indefinite integrals

Indefinite Integrals

Indefinite integrals

indefinite integrals

Indefinite integrals

Search Tags